Question

Find the equation of the ellipse whose foci are $(4,6)$ & $(16,6)$ and whose semi-minor axis is $4.$

• A. $\frac{(x-6{)}^2}{26}+\frac{(y-10{)}^2}{16}=1$
• B. $\frac{(x-10{)}^2}{26}+\frac{(y-6{)}^2}{16}=1$
• C. $\frac{(x-6{)}^2}{52}+\frac{(y-10{)}^2}{16}=1$
• D. $\frac{(x-10{)}^2}{52}+\frac{(y-6{)}^2}{16}=1$

Answer 1

The correct option is D $\frac{(x-10{)}^2}{52}+\frac{(y-6{)}^2}{16}=1$

Semi minor axis length $b=4$
Center of ellipse is midpoint of foci$=(\frac{4+16}{2},\frac{6+6}{2})=(10,6)$
Distance between focus and center $ae=10-4=6$
$a^2{e}^2=36=>a^2(1-\frac{b^2}{a^2})=36=>a^2-b^2=36=>a^2=36+4^2=52$
Therefore equation of ellipse is $\frac{(x-10{)}^2}{52}+\frac{(y-6{)}^2}{16}=1$